Stitz-Zeager_College_Algebra_e-book

# Moving on to the case where r r we have that 2k

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Unformatted text preview: rawn below. β ≈ 11.81◦ c=4 β ≈ 108.19◦ a=3 α = 30◦ γ ≈ 41.81◦ α = 30◦ c=4 a=3 γ ≈ 138.19◦ b ≈ 5.70 b ≈ 1.23 ◦ 6. For this last problem, we repeat the usual Law of Sines routine to ﬁnd that sin(γ ) = sin(30 ) so 4 4 that sin(γ ) = 1 . Since γ must inhabit a triangle with α = 30◦ , we must have 0◦ < γ < 150◦ . 2 Since the measure of γ must be strictly less than 150◦ , there is just one angle which satisﬁes both required conditions, namely γ = ◦ ◦ . So β = 180◦ − 30◦ − 30◦ = 120◦ and, using the 30 √ sin(120 Law of Sines one last time, b = 4 sin(30◦ ) ) = 4 3 ≈ 6.93 units. c=4 β = 120◦ a=4 γ = 30◦ α = 30◦ b ≈ 6.93 Some remarks about Example 11.2.2 are in order. We ﬁrst note that if we are given the measures of two of the angles in a triangle, say α and β , the measure of the third angle γ is uniquely determined using the equation γ = 180◦ − α − β . Knowing the measures of all three angles of a triangle completely determ...
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